Solve (sqrt{7/6}-3sqrt{43})*sqrt{3}

$$( \sqrt { \frac { 7 } { 6 } } - 3 \sqrt { 43 } ) \times \sqrt { 3 }$$

Evaluate

$$\frac{\sqrt{14}}{2}-3\sqrt{129}\approx -32.202621381$$

Short Solution Steps

Rewrite the square root of the division \sqrt{\frac{7}{6}} as the division of square roots \frac{\sqrt{7}}{\sqrt{6}}.

$$\left(\frac{\sqrt{7}}{\sqrt{6}}-3\sqrt{43}\right)\sqrt{3}$$

Rationalize the denominator of \frac{\sqrt{7}}{\sqrt{6}} by multiplying numerator and denominator by \sqrt{6}.

$$\left(\frac{\sqrt{7}\sqrt{6}}{\left(\sqrt{6}\right)^{2}}-3\sqrt{43}\right)\sqrt{3}$$

The square of \sqrt{6} is 6.

$$\left(\frac{\sqrt{7}\sqrt{6}}{6}-3\sqrt{43}\right)\sqrt{3}$$

To multiply \sqrt{7} and \sqrt{6}, multiply the numbers under the square root.

$$\left(\frac{\sqrt{42}}{6}-3\sqrt{43}\right)\sqrt{3}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -3\sqrt{43} times \frac{6}{6}.

$$\left(\frac{\sqrt{42}}{6}+\frac{6\left(-3\right)\sqrt{43}}{6}\right)\sqrt{3}$$

Since \frac{\sqrt{42}}{6} and \frac{6\left(-3\right)\sqrt{43}}{6} have the same denominator, add them by adding their numerators.

$$\frac{\sqrt{42}+6\left(-3\right)\sqrt{43}}{6}\sqrt{3}$$

Do the multiplications in \sqrt{42}+6\left(-3\right)\sqrt{43}.

$$\frac{\sqrt{42}-18\sqrt{43}}{6}\sqrt{3}$$

Express \frac{\sqrt{42}-18\sqrt{43}}{6}\sqrt{3} as a single fraction.

$$\frac{\left(\sqrt{42}-18\sqrt{43}\right)\sqrt{3}}{6}$$

Use the distributive property to multiply \sqrt{42}-18\sqrt{43} by \sqrt{3}.

$$\frac{\sqrt{42}\sqrt{3}-18\sqrt{43}\sqrt{3}}{6}$$

Factor 42=3\times 14. Rewrite the square root of the product \sqrt{3\times 14} as the product of square roots \sqrt{3}\sqrt{14}.

$$\frac{\sqrt{3}\sqrt{14}\sqrt{3}-18\sqrt{43}\sqrt{3}}{6}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{3\sqrt{14}-18\sqrt{43}\sqrt{3}}{6}$$

To multiply \sqrt{43} and \sqrt{3}, multiply the numbers under the square root.

$$\frac{3\sqrt{14}-18\sqrt{129}}{6}$$